(D) The Mean Value Theorem $(MVT)$ requires a function to be continuous on $[a, b]$ and differentiable on $(a, b)$.For option $D$,$f(x) = |x|$ is not

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(D) The Mean Value Theorem $(MVT)$ requires a function to be continuous on $[a, b]$ and differentiable on $(a, b)$.For option $D$,$f(x) = |x|$ is not differentiable at $x = 0$. Since $0 \in [0, 1]$,the function $f(x) = |x|$ is not differentiable on the interval $(0, 1)$.Therefore,the Mean Value Theorem cannot be applied to $f(x) = |x|$ on $[0, 1]$ because it fails the differentiability condition at $x = 0$.

Class 12 Mathematics Continuity and Differentiation Rolle’s theorem, Lagrange's mean value theorem

Medium

According to the Mean Value Theorem,which of the following functions does not satisfy the conditions on the interval $[0, 1]$?

A
$f(x) = \begin{cases} \frac{1}{2} - x, & x < \frac{1}{2} \\ (\frac{1}{2} - x)^2, & x \ge \frac{1}{2} \end{cases}$
B
$f(x) = \begin{cases} \frac{\sin x}{x}, & x \ne 0 \\ 1, & x = 0 \end{cases}$
C
$f(x) = x |x|$
D
$f(x) = |x|$

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Rolle’s theorem, Lagrange's mean value theorem

The equation $2^x+5^x=3^x+4^x$ has

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If the function $f(x) = x^3 - 6x^2 + ax + b$ satisfies Rolle's theorem in the interval $[1, 3]$ and $f'\left( \frac{2\sqrt{3} + 1}{\sqrt{3}} \right) = 0$,then $a = $ ..............

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Let $f$ be a function that is derivable on the interval $[0, 1]$. Then, which of the following statements is true?

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Let $f(x)$ be continuous on $[0, 5]$ and differentiable in $(0, 5)$. If $f(0) = 0$ and $|f^{\prime}(x)| \leq \frac{1}{5}$ for all $x$ in $(0, 5)$, then which of the following is true for all $x$ in $[0, 5]$?

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According to the Mean Value Theorem,which of the following functions does not satisfy the conditions on the interval $[0, 1]$?

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The equation $2^x+5^x=3^x+4^x$ has

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